Divide the Sequence

Divide the Sequence

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1462    Accepted Submission(s): 693

Problem Description

Alice has a sequence A, She wants to split A into as much as possible continuous subsequences, satisfying that for each subsequence, every its prefix sum is not small than 0.

 

Input

The input consists of multiple test cases.
Each test case begin with an integer n in a single line.
The next line contains n integers A1,A2⋯An .
1≤n≤1e6
−10000≤A[i]≤10000
You can assume that there is at least one solution.

 

Output

For each test case, output an integer indicates the maximum number of sequence division.

 

Sample Input

  

   6
1 2 3 4 5 6
4
1 2 -3 0
5
0 0 0 0 0
  

 

Sample Output

  

   6
2
5
  

 

Author

ZSTU

 

Source

2016 Multi-University Training Contest 5

 

题意分析：把长度为n的序列分成尽量多的连续段，使得每一段的每个前缀和都不小于0。保证有解。


从后往前贪心分段即可。大于等于0的为一段，遇到负数就一直相加到非负为止！（注意精度问题 用long long）

AC代码：

#include<iostream>
#include<algorithm>
#include<stdio.h>
using namespace std;
typedef long long LL;
const int N=1000010;
int a[N];
int main()
{
        LL n,i;
        while(scanf("%lld",&n)!=EOF)
        {
                for(i=1;i<=n;i++)
                        scanf("%lld",&a[i]);
                LL sum=0;
                for(i=n;i>=1;i--)
                {
                        sum+=a[i];
                        if(sum>=0)
                                sum=0;
                        else n--;
                }
                printf("%lld\n",n);
        }
        return 0;
}